POW(x,n) | Binary Exponentiation | Leetcode

[@takeUforward]

Understood or not ? Drop a Yes or No :D .. . Check me out at Instagram: instagram.com/striver_79/ . . If you appreciate the channel's work, you can join the family: bit.ly/joinFamily

[@abhiprajapati2835]

Make videos faster

[@sahilchoudhary1473]

understood

[@mutyamreddybejjanki8307]

Great job bro 🙌

[@umapandey6796]

yes

[@mohammedwaseem8599]

Yes.

[@rishitiwari6200]

Bro don't end this series and please ignore dislikers and don't get demotivated as I came to know about many technique to solve same problem which I submitted in brute force approach

[@abhishekkmandal]

sound quality has exponentially improved :)

[@iampatelajeet]

Raj bhai your placement series is really really useful for all students thanks for your efforts .

[@sudeshpawar733]

I knew this approach but it's always better to understand it from the legend. thanks for making our life easy bro ❤️

[@takeUforward]

long wla pta tha ?

[@sudeshpawar733]

@takeuforward Nahi Bro.. as you said this video has something for everyone.

[@harshitsaxena7205]

@@takeUforward that's why striver is "THE STRIVER"

[@himanshumishra8740]

The overflow problem can be solved without using long and just writing n=abs(n). I did this and it worked fine for me.

[@rohandeshmukh3989]

but i tried to do the way striver did that multiplying it by -1 but it was giving me the error that it cannot store that much value in int and then instead of using that i used abs and it worked but i didn't get why does it works it is same as the first approach i used but still i got wa on first approach

[@visheshkaran7083]

@@rohandeshmukh3989 bro typecast the right hand side with (long) it won't give you error

[@satyamsrivastava.]

Nice... It is called as Fast Exponential Algorithm I had studied it during B.Tech

[@Rajesh-op8zx]

you're from which college coz even our IITs dont tech DS ,Algo to us

[@aakarsrivastava7193]

Yeah man, I'm in IIT doing final year of B.Tech+M.Tech in Computer Science and still I've to come to this video to teach myself.

[@satyamsrivastava.]

So I was preparing for Gate during B.tech at that time, I studied this in Algorithms subject by myself. Same my college didn't teach me this.

[@vibhor3049]

Using recursion- JAVA static double pow(double x, long n){ if(n == 0) return 1; if(n%2 == 0) return pow(x*x, n/2); else return x*pow(x, n-1); } When the function returns the result, do check if n was +ve or -ve and show output accordingly.

[@bisatisrilatha7957]

why use double here and why long for n?????????

[@democratcobra]

Negative Handel hota hein keya/?

[@rajrangwani5669]

Very nice video...already knew the logn solution but the overflow condition was new which could've easily ruined the interview.......please continue on this series it helps a lot

[@nilesh69420]

Congrats Sir. Big fan.

[@2018-y8t]

Keep going! This is really helpful!

[@aaryannigam9552]

I think there is some fault in c++ solution, the if condition is wrong

[@studyacc1234]

you didn't explain the recursive one..

[@AnkitRaj-ux7lv]

//correct code best excpected by any company class Solution { public: double myPow(double x, long long n) { double ankit= pow(x,n)1.0; if(n

[@paritoshlambat656]

Do u have any idea bitwise operator use that

[@divaynagar1929]

Can I also do the bitwise operators approach , that also works in O(logn) as the number of bits in a number is O(logn) , Please Tell can I do it for that also ??

[@ravishankar9247]

Why have you kept this question in the ARRAYS section of the SDE sheet ??

[@kaustavpaul9306]

Brute Force class Solution { public double myPow(double x, int n) { double ans = x; if(n > 0){ for(int i=1; in; i--) ans = ans / x; } return ans; } }

[@syagni_]

dont know why, but even after copy-pasting your code i am getting "time limit exceeded" in submission :/

[@kaustavpaul9306]

@@syagni_ yaa thats y its Brute force

[@amitagrawal4660]

All cases got passed except the one below: x = 2.00000 n = -2147483648 Output = 1.00000 Expected = 0.00000

[@deepakchoudhary6593]

use long num= n; because Integer.MAX_VALUE is 2147483647 and Integer.MIN_VALUE is -2147483648 and when you convert to positive it overflows for Int. Hope you understood.

[@AnkitSingh-ut7et]

@@deepakchoudhary6593 but the idea is not to use trick and compute answer but to find some generic solution to handle such edge case.

[@yatinarora1252]

Hey u can use it in a edge case if n>INT_MIN return 0.0000

[@mutyamreddybejjanki8307]

Don't dislike , he is doing a great job ❤ thank you bhai 😎🙌

[@AnkitKumar-ix3fl]

This is what we do in binary exponentiation. Bro your videos is very helpful. Keep making this type of content and if u can please make series on dp.

[@gamingwithhemend9890]

Where is this series in your channel. Please share the playlist link (placement series) here

[@deep90402]

Why we r making power +ve when it is negative??

[@priyankavasam8769]

this logic is just mind blowing.. I wasn't able to think in that angle

[@vishwa5829]

better explain using recursion tree.easy to understand

[@AdityaSharma-er3gs]

Can't we just use in build pow() ?

[@aryamanjaiswal5163]

Hello bhaiya, can i know y r u doing this=> ans = (double)(1.0)/(double)(ans), ans and 1.0 are already double then y type casting required?

[@shivampurbia6169]

Why this method and recursion is taking 20ms but inbuilt pow() or ** operator is taking 12 ms why the later one is fast ?

[@arnabbiswas5472]

I think the solution will not work for n=INT_MIN.

[@alokmishra8786]

thank you for the best explanation and this is very helpful

[@MsKoniki]

please keep posting bro . really need your help to get placed. still i am begginer into competeive coding

[@aysams2]

Hey, did you get placed? What company do you work in now?

[@MsKoniki]

@@aysams2 Thankfully yes , i got placed in serviced based MNC and switched to a fintech after that .

[@adarshrai9516]

Thanks for everything ❤️❤️❤️❤️

[@kanishkanim]

Appreciable....your videos are helping a lot....carry on!!! 🤩🤩

[@ranasauravsingh]

UNDERSTOOD... !!! Thanks striver for the video... :)

[@rohitrai2133]

why there are so many dislikes in leetcode for this problem!!

[@alexmercerind]

Thanks! I was using a recursive approach before. I wonder how to handle mod now. EDIT: Nevermind. Thanks for great content.

[@govindsharma7696]

Bhaiya aapse ek request h aap apni 1st year se 4th year / placement tak ki journey share kijiye .. Aap bhi batana ki aapne kya kya problem face ki starting m aapne kese kese overcome kiya ... Kya kya resources use kiye .. and so on .. I know ki sabki journey alag h and har koi ek hi jesi journey nhi follow krta but hume thoda idea lagega ki hume kitni mehnat ki jarurat h tier 3 se achi company tak jane k liye ... Pls do this na bhaiya it's a humble request to u

[@anveshasharma2726]

Shall we do this with recursion in interview?

[@lelouchlemprouge6380]

Math.exp(n*Math.log(x)); Ye karne se ajaega aur x positive negative ka sign bit save krdena i think this is fastest?

[@AkashSingh-mk7od]

Very nicely explained.... it's clearly understood....thankyou bhaiya....your series can definitely help someone feel confident and work more hard.

[@roshni9320]

This solution is still giving me TLE for the long long test case!

[@ShreyaSingh-vr9qi]

Nice explaination !! bs bhaai videos rokna nahi puri 180 days complete karna.. Thanks so much for helping..

[@blitzkrieg5454]

Hats off to all your hardwork, keep going!

[@abhinavmishra7617]

Although the complexity of both approaches is identical, this approach will be faster in practice since we have the overhead of the recursive calls.

[@DR-mq1le]

in the else condition should it not be "ans=x*x" ?

[@amansinghal4663]

My recursive solution ( accepted in leetcode ) class Solution { public: double func(double x, int n){ if(n==0){ return 1; } if(n%2 == 0){ return func(x*x, n/2); } else{ return x*func(x*x, n/2); } } double myPow(double x, int n) { if(n

[@neelesh621]

the fun faCT is, there is no difference between c++ & java solution

[@davegupta329]

double myPow(double x, int n) { double pow; pow = exp(n * log(fabs(x))); if (x < 0 && n % 2 != 0) pow = -1 * pow; return pow; } Can it be considered?

[@mohammedwaseem8599]

This code is giving Wrong Answer.

[@takeUforward]

The code works fine..

[@deepakavva317]

I think a series on dynamic programming can increase your subscribers.

[@ramsai6141]

yes brother...so many people are waiting for your DP series

[@devangpapinwar208]

Thank you Trevor

[@Morimove]

this video is linked to recursion on DSA A2Z series but the shown solution is iterative

[@takeUforward]

Yes because this can also be solved using recursion. I will soon add

[@Morimove]

@@takeUforward thankyou sir

[@amd7407]

bro awesome videos ,your sheet helps me a lot , little sound can be improved . But please keep making videos .

[@raholsaha8191]

Nice work bro. You are doing a great job.

[@pleasesirmorevideos4684]

Why dislikes???

[@DR-mq1le]

in the article the c++ code is wrong ,

[@chandlerbing8164]

not accepted by leetcode submission please explain floating point power as well brother

[@himanshidixit6406]

Completely understood..Thank you so much..:)

[@samqureshi9415]

First time dekha video isse phoe ds or algo kya hai pata he nahi tha . Per dekh kar acha laga ❤️

[@mdmuquimakhter5145]

class Solution { public: double myPow(double x, int n) { long long num=n; double ans=1.0; if(num

[@wisdomkhan]

how is this interview specific?

[@omanshsharma6796]

The quality of explanations seems to be inversely related to the quality of mic :D

[@swetantkumar7029]

Following ur series and waiting next video to come eagerly....Great job man.

[@sanskarghyar9408]

class Solution { public: double myPow(double x, int n) { if(n

[@Techna-to5qt]

what if n is a real number?

[@SHIVAMGUPTA-wb5mw]

Following ur series ...plz continue

[@kaichang8186]

understood, thanks for the awesome video

[@WasimAkram-hb5lr]

Brother your is really helpful keep going and ignore hate comments

[@akashnarvaria3663]

Why we have to take long nn? With int nn it is not working Can someone explain?

[@shreyash8273]

Int will overflow for the -intMin value when will make it positive

[@deepapandey849]

Amazing series + solutions, hats off! Just, this video has very annoying sound quality :(

[@chakshusalgotra]

thanks all to u sir u been a great help

[@parikshitrathore1510]

In some places the voice of previous and succeding transitions are overlapping!!

[@parikshitrathore1510]

But, understood!!

[@takeUforward]

@@parikshitrathore1510 Yeah am using a new editor hence this is happening, will check this bro in the next one !

[@dhruvrajput5194]

Really great explanation.. But i think there is some problem with your mic.. Or maybe something's wrong with my earphones😕

[@takeUforward]

Some problem on my side!

[@ketanahuja8939]

bhiaya not got this line x = x* x; and we are doing nn/=2;

[@linhinNTU]

can anyone explain for me why n needs to be " long long" type?

[@jiviteshvarshney3644]

For saving large int

[@sheikhaman6218]

Always a helping hand... 😇

[@Rahul-tk8ck]

U r helping me alot bro.stay safe✨

[@sumitkulyal4102]

Good work bhaiyya. Haters are doing great job, bhaiyya it's means your are doing great job more haters means you are doing right work. Keep going bhaiyya.

[@samarthajadhao1176]

can make videos on C++ tricks...

[@sarthak1317]

The voiceover at 2:22 xD

[@letsexplore7590]

yes

[@yashkhatwani3198]

Thank you

[@lakshkaushik6366]

Can u help me by guiding me how to start doing this stuff I am in 12 th now

[@39_jatinjain4]

can anyone plz tell me where i am doing wrong?? class Solution { public: double sol(double x,long n){ // base case if(n==0) return 1; return x*sol(x,n-1); } double myPow(double x, int n) { long z=abs(n); if(n%2!=0) z=z+1; double t= sol(x,z/2); double ans=t*t; if(n%2!=0) ans=ans/x; if(n

[@RaaheeAmit]

U help us alot

[@divyanshigupta1568]

You are truly a legend 👏🏻

[@nopecharon]

Thanks

[@AshishKumar-me8pr]

Stiver_79 you are Genius.. excellent

[@redhathacker3458]

Yes

[@atulmalakar]

Naming of variables could have been better.

[@BRSanush]

Happy teachers day Striver sir🎉

[@het314]

using long is kinda cheating. Try diving when n

[@israkulhaquetusher3256]

SDE sheet. link isn't working

[@tannukumari5144]

great work bhaiya, thank you

[@abhinavmishra7617]

why didn't we do binary exponentiation? fast_power.... pow(a,n){ base cases b=pow(a,n/2); if(n%2==1) return b*b*a; return b*b; }

[@takeUforward]

Watch the video, you wrote the recursive solution for the same. Dont learn the codes, understand the algo, then you will know that both are same!

[@abhinavmishra7617]

@@takeUforward bhai maine yaad to nahi kiya h....ye kaise bol diye aap? n if mai apne doubt clear krne k liye comment krunga to tum aise reply karoge? not cool

[@takeUforward]

Nai bro, aapne pucha binary exponentiation kyu use nai kia aapne. Yahi comment tha? Th mne suggestion dia. Kyu ki agar algos smjhoge th ye puchte nai. Baaki tum isse constructive loge toh next time sayad tm analysis kroge :) rest me chahta th comment ka ni v reply kr skta tha.. but me kia so that tm smjho ye chiz ki tm algos smjhe ni. Logic tmhre code ki smjhte th ye question aati hi nai as dono same. Bs video wali iterative h tumhri code recursive!!

[@abhinavmishra7617]

@@takeUforward ok bhai... actually mere liye recursive zyada intuitive h maybe isliye main iteratively nhi smjh paya isliye bs maine puchha

[@takeUforward]

Abhinav Mishra intuitive ho sakta hai but agar tumhe pta h n odd p n-1 kr rhe and n even pe n/2, toh ye recursion me kyu kr rhe ho? Wahi reason mene bataya h. Th dono same hua na ?? Interviews m agar tum ase dono ko alag bologe th dikkat ho skti, iskie ache se analyse kro, like try to write both iterative and recursive

[@Learnprogramming-q7f]

Thank you Bhaiya

[@enigma2886]

thanks !

[@MaheshPatil-of1zy]

its 2024 feb i am watching 😊

[@aditya_sipani]

what is nn here,like ?

[@Ankitkumar-gc5cc]

That is just storing value of n so that while we change the value of nn in loop we can retain value of n and that can be used for final return statement where he is checking if it's negative or positive

[@tejasjoshi1907]

Why this question has so many dislikes on leetcode.??

[@gauravidesigns]

Thankyou bro